EVEN06 - PART II CHAPTER 6 Applications of Integration...

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Unformatted text preview: PART II CHAPTER 6 Applications of Integration Section 6.1 Section 6.2 Section 6.3 Section 6.4 Section 6.5 Section 6.6 Section 6.7 Area of a Region Between Two Curves . . . . . . . . . . 264 Volume: The Disk Method . . . . . . . . . . . . . . . . . 271 Volume: The Shell Method . . . . . . . . . . . . . . . . 278 Arc Length and Surfaces of Revolution . . . . . . . . . . 282 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287 Moments, Centers of Mass, and Centroids . . . . . . . . . 290 Fluid Pressure and Fluid Force . . . . . . . . . . . . . . . 297 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299 Review Exercises Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305 CHAPTER 6 Applications of Integration Section 6.1 Area of a Region Between Two Curves Solutions to Even-Numbered Exercises 2 2 2. A 2 2x 5 x2 2x 1 dx 2 x2 4 dx 1 1 4. A 0 x2 x3 dx 6. A 2 0 x 1 3 x 1 dx 1 3 8. 1 1 x2 y x2 1 dx 10. 2 y 7 x3 3 x x dx 3 4 12. 4 sec2 x y cos x dx 2 6 5 4 3 x 2 (3, 6) 3 ( (2, 2 ) 3 2 −π,2 4 2 ) π ( 4 , 2) −2 2 1 −1 (3, 1) x 4 5 6 7 π (− 4 , 22 ) (π , 22 ) 4 −π 4 −2 π 4 x 8 14. f x gx A 1 2 2 1 2x 16. A 2 10 8 2 1 x 2 7 x 2 3 xx 8 10 dx 8 8 dx x 32 x 8 7x2 4 112 Matches (a) y x3 8 64 10x 2 80 1 7 20 18 4 y 3 (2, 9) (0, 2) 8 6 (8, 6) 1 (4, 0) x 1 2 3 4 2 (2, 9 ) 2 (8, 0) x 2 4 6 10 264 S ection 6.1 18. The points of intersection are given by x2 4x x2 1 3x x2 3 Area of a Region Between Two Curves 265 20. The points of intersection are given by: x2 4x x3 2 x x 2 0, 3 x 0 1 0 when x 3x when x 1 x 0, 3 A 1 dx 3 fx 0 3 g x dx 4x 3x d x 2 x x3 3 2 dx 32 x 2 3 0 A 0 3 x2 x2 0 4x 3x dx 3x 2 23 0 x2 0 3 x 3 9 y 6 5 4 3 2 3 x2 0 y 6 5 4 3 9 2 27 2 9 2 (3, 5) (3, 4) (0, 2) (0, 1) x 1 2 3 4 5 6 −1 x 1 2 3 4 5 5 22. A 1 y 1.0 0.8 0.6 0.4 0.2 1 x2 (1, 1) 0 dx 1 x 5 1 4 5 (5, 0.04) x 1 2 3 4 5 24. The points of intersection are given by 3 x x 1 1 2x 2 1 x x 0 1 1 3 y x3 3x2 3x 1 1 (2, 1) (1, 0) x 2 x3 x x2 xx 1 3x 2 3x 2x x 1 x 1 0 0⇒x 3 0, 1, 2 (0, −1) A 2 0 x 1 1 dx 1 43 0 2 2 x2 2 1 2 3 x 4 0 3 4 1 2 266 Chapter 6 Applications of Integration 26. The points of intersection are given by: 2y yy 3 y2 3 fy y (−3, 3) y 3 0 when y g y dy 0, 3 A 0 3 1 (0, 0) 2y 0 3 y2 y2 dy y dy 32 y 2 13 y 3 3 0 −3 −2 −1 x 1 3y 0 9 2 3 y 28. A 0 3 0 fy y 16 3 g y dy 4 3 y2 y2 3 0 dy 12 ( 3 ,3 7 ) 2 1 2 16 0 2y d y 7 1.354 −1 1 x 1 2 3 16 y2 0 4 1 30. A 0 4 4x 4 4 2 4 ln 2 x y dx (1, 4) 1 x 0 3 (0, 2) 4 ln 2 1 −1 x 1 3 1.227 32. The point of intersection is given by: x3 2x x3 1 34. The points of intersection are given by: x4 1 A x2 x2 2 1 1 2x 0 when x g x dx 2x 2 4 2x 2 0 2 2x 2 0 when x x4 x4 dx x5 5 2 0 0, ± 2 A 1 1 fx x3 1 1 2 2 0 2x 2 d x 2x 1 dx 1 x4 4 2x d x 1 4x 2 4x 3 3 x 1 3 x 1 2 2 128 15 Numerical Approximation: 2.0 y Numerical Approximation: 8.533 10 (−2, 8) (2, 8) (−1, 2) −4 4 −2 (1, 0) −2 −1 −1 −2 x 2 (0, 0) (1, −2) S ection 6.1 36. f x x4 4x 2, g x x3 4x Area of a Region Between Two Curves y 267 The points of intersection are given by: x4 x4 xx 0 g4 3 4x 2 4x 2 x3 0 4x −4 −3 −1 x 1 3 4 x3 1x 4x 2 2x 0 when x 1 2, 0, 1, 2 2 −3 −4 f A 2 x3 248 30 37 60 4x 53 60 x4 293 30 4x 2 d x 0 x4 4x 2 x3 4x d x 1 x3 4x x4 4x2 d x Numerical Approximation: 8.267 3 0.617 0.883 9.767 4 38. A 0 6x x2 1 1 0 dx 3 3 40. A x 0 2 ( 3, ) 0 −1 −1 9 5 4 4 x dx x 3.434 3 ln x 2 3 ln 10 6.908 (0, 0) 5 −1 5 Numerical Approximation: 6.908 6 −1 1 42. A 2 cos 2x 1 sin 2x 2 3 4 y sin x dx 6 44. A 0 2 2 42 x 2 4x 4x 4 4 1 4 4 4 sec sec x x tan dx 4 4 x 4 1 cos x 2 0 3 2 0 33 4 1.299 2 2 y 4 3 2 1 24 2 2 2 2 4 ( −π 2 π, 1 62 ) x 2.1797 π 6 −1 (− π , −1) 2 (1, 2) x 1 46. From the graph we see that f and g intersect twice at x and x 1. 1 y 0 48. A 0 2 sin x 2 cos x cos 2x 1 sin 2x 2 (π, 1) 5 4 0 dx 4 0 A 0 1 gx 2x 0 f x dx 3 (1, 3) 2 1 3x dx 1 0 −1 (0, 1) 2 − x2 21 x 1x 3 ln 3 (0, 1) 4 x 1 2 −2 1 ln 3 0.180 268 Chapter 6 5 Applications of Integration 50. A 1 4 ln x x 5 2 1 0 dx 2 ln 5 2 52. (a) y 5.181 (b) A 0 x e x, y 1 0, x 0, x 1 2 ln x 2 x e x d x. No, it cannot be evaluated by hand. (5, 1.29) (c) 1.2556 6 4 y = xe x 0 (1, 0) −2 −1 −1 2 x 54. F x 0 12 t 2 0 2 dt 13 t 6 x 2t 0 x3 6 2x (b) F 4 y (a) F 0 y 43 6 24 56 3 (c) F 6 y 36 12 48 20 20 16 16 12 12 8 8 4 4 x 1 2 3 4 5 6 1 2 3 4 5 6 x 20 16 12 8 4 x 1 2 3 4 5 6 y y 56. F y 1 4ex 2 dx 1 y 30 25 20 15 10 5 −1 8ex 2 1 8e y 2 8e 12 (a) F 0 (b) F 0 y 30 25 20 15 10 5 x 8 8e 12 3.1478 (c) F 4 y 30 25 20 15 10 5 8e2 8e 12 54.2602 1 2 3 4 −1 x 1 2 3 4 −1 x 1 2 3 4 4 58. A 2 4 2 9 x 2 7 x 2 6 12 7 dx x 6 4 5 dx 4 5 x 2 16 x 5 dx 7 x 2 21x 21 d x 6 72 x 4 y 6 4 2 4 7x 2 72 x 4 7 4 7 14 9 y = 2 x − 12 (4, 6) 5 y = − 2 x + 16 (6, 1) x 2 −2 −4 6 8 10 (2, −3) y=x−5 S ection 6.1 1 x2 x2 1 2x 1 2 Area of a Region Between Two Curves y 269 60. f x fx (0, 1) 3 4 1 2 1 4 1 f ( x) = 2 x +1 1 At 1, , f 1 2 Tangent line: y 1 2 1 x 2 1 . 2 1 x 2 1 x2 1 1 dx (1, 21 ) y=− 1x+1 2 x 1 2 1 3 2 2 1 or y 1 at x 0. arctan x x2 4 1 The tangent line intersects f x 1 A 0 1 x2 1 1 x 2 x 0 3 4 0.0354 62. Answers will vary. See page 417. 64. x 3 ≥ x on 1, 0 1 y x 3 ≤ x on 0, 1 Both functions symmetric to origin 0 1 (1, 1) (0, 0) x3 1 1 x dx 0 x3 0. 2 x2 2 x d x. −1 x 1 −1 Thus, 1 x3 1 x dx x (− 1, − 1) A 2 0 x3 dx x4 4 1 0 1 2 66. Proposal 2 is better, since the cummulative deficit (the area under the curve) is less. 9 68. A 2 0 9 2 x dx 9 0 9 b b 2 9x 9 9 x b bx 9 x2 2 b9 9 b 9 x2 2 9 y 81 0 12 9 b dx x dx 9 0 b 81 2 81 2 81 2 81 2 9 2 2.636 (− (9 − b), b) 6 (9 −b, b) x 3 6 2 0 −6 −3 −3 −6 29 b b 9 2 270 Chapter 6 n Applications of Integration y 70. lim →0 i 4 1 x i2 2 x 4i and x n 4x 4 is the same as n x3 3 2 2 5 f ( x) = 4 − x 2 where xi 2 3 2 4 2 x2 dx 32 . 3 (− 2, 0) −3 −1 1 1 −1 (2, 0) x 3 5 5 72. 0 7.21 0.26t 0.02t 2 7.21 0.1t 0.01t 2 d t 0 0.01t 2 0.01t 3 3 29 billion 12 0.16t d t 0.16t 2 2 5 0 $ 2.417 billion 74. 5% : P1 31 %: P2 2 5 893,000 e 0.05 t 893,000 e 0.035 t Difference in profits over 5 years: 893,000e 0.05t 0 893,000e 0.035t d t 893,000 e 0.05t 0.05 e 0.035t 0.035 5 0 893,000 25.6805 893,000 0.2163 Note: Using a graphing utility you obtain $193,183. 34.0356 $ 193,156 20 28.5714 76. The curves intersect at the point where the slope of y2 equals that of y1, 1. y2 0.08x2 k ⇒ y2 0.16x 1⇒x 1 .16 6.25 6.25 (a) The value of k is given by y1 6.25 k y2 0.08 6.25 3.125. 2 (b) Area k 2 0 6.25 y2 y1 d x 3.125 x2 2 x dx 6.25 0 2 0 0.08x 2 0.08x 3 3 2 3.125x 2 6.510417 1 16 2 13.02083 78. (a) A (b) V 80. True 6.031 2A 2 2 5.908 2 11.816 m 3 1 8 2 5.908 (c) 5000V 5000 11.816 59,082 pounds Section 6.2 Volume: The Disk Method 271 Section 6.2 2 Volume: The Disk Method 2 2. V 0 3 4 x2 2 dx 0 3 x4 8x 2 16 dx x5 5 8x 3 3 2 16x 0 256 15 x2 4 x2 2 2 4. V 0 9 x2 2 dx 0 9 9x x2 dx x3 3 3 6. 2 18 8 x2 x 4 16 8 ±2 V 2 2 2 2 4 2 0 x2 4 2 2 2 dx 0 x4 16 2x3 3 2x2 12x 12 dx 2 0 2 2 2 2 x5 80 128 2 80 32 2 3 132.69 24 2 448 2 15 x2 ⇒ x 16 0 4 4 8. y V 16 4 16 2 y2 4 10. V 1 y2 y5 5 459 15 2y 4 153 5 4y 2 d y 1 y4 4 1 8y 3 16y 2 d y y2 4 0 dy 0 16 y2 dy 16y 3 3 16y y3 3 0, x 2, r y 8 128 3 2 y2 y dy 2 4y y2 4 8 12. y 2x 2, y (a) R y V (b) R x 16 0 2x 2, r x 2 0 4x 5 5 2 0 4 0 y 8 6 V 0 4x 4 d x y 8 6 128 5 4 2 x 2 4 −4 −2 4 2 x 2 4 −4 −2 (c) R x V 8, r x 2 8 64 4x 4 15 x 5 2 0 2x 2 32 x 2 dx 4 0 (d) R y 4 x4 dx 2 2 8 y 2, r y 2 y 2 4 y 2 2 0 dy y dy 2 y2 4 8 0 64 0 2 V 0 8 32x 2 0 8x 2 x4 dx 4 0 4 83 x 3 y 896 15 4y y 8 4 2 32 y 3 16 3 6 6 4 2 x 2 4 4 2 x 4 −4 −2 −4 −2 272 14. y Chapter 6 6 2x 6 0 Applications of Integration x 2, y 2x 6 x 6 intersect at x 2 3, 3 and 0, 6 . (b) R x 6 0 (a) R x V x 2, r x 2x 4x 3 x4 y 8 6 6 2 2x 3 2x x2 x2 3, r x 2 x 3 2 6 dx 3 x2 x dx V 3 0 x 3 0 x4 3 9x 2 3x 3 36x d x 3 0 x4 243 5 15 x 5 4x 3 x4 y 8 3x 2 x3 9x 2 18x d x 0 3 15 x 5 18x 2 3 108 5 4 2 x 2 4 2 x 2 −6 −4 −2 −6 −4 −2 16. R x 4 2 x3 ,r x 2 4 x3 2 2 0 18. R x V 4, r x 3 4 2 sec x 4 sec x 2 4 0 3 dx V 0 2 dx x6 4 x7 28 128 28 2 0 16 0 4x3 x4 16 8 sec x dx 0 sec 2 x d x 3 16x 32 y 8 ln sec x 8 ln 2 8 ln 2 y 5 tan x 3 3 3 3 tan x 0 144 7 8 ln 1 27.66 0 0 (2, 4) 3 2 1 −1 −1 x 1 2 3 4 3 2 1 π 9 π 3 x 2π 9 4π 9 5π 9 20. R y V 6, r y 4 6 y y3 3 4 0 2 6 dy 368 3 y y 5 4 3 2 1 y 6 0 2 36y x 1 −1 2 3 4 5 S ection 6.2 6 ,ry y 6 2 6 Volume: The Disk Method 273 22. R y V 36 6 6 0 24. R x V 2 x4 2 x 2, x2 rx 2 0 6 y 1 2 dy 2 y 1 dy y2 1 y 6 2 x4 0 2 dx 2 0 4x 2 4x 3 3 x4 dx x5 5 2 0 2 36 36 36 y 6 5 4 y 35 6 13 3 2 ln y 2 ln 6 1 2 ln 3 2 3 2 128 15 2 ln 2 6 ln 3 241.59 y 12 13 3 2 1 −3 −1 x 1 2 3 −2 3 −3 2 1 x 1 2 3 4 5 26. R x V 3 x 8 0 1 x 8 , rx 2 0 dx 2 28. R x V e x 2, r x 4 0 dx 3 1 1 1 x 1 ex 0 4 22 9 0 x dx 8 ex dx 0 4 8 0 9 y 4 ex 0 e4 y 8 1 168.38 3 6 2 4 1 2 x 2 4 6 8 −2 x 2 4 6 4 30. V 0 4 0 4 x2 4 1 x 2 5x 5x2 2 2 x 16 dx 2 8 dx 4 8 4 x x2 4 5x2 2 5x 2 4 16 dx 8 1 x 2 2 dx 4 3 2 y (4, 2) x3 12 88 3 4 16x 0 x3 12 16x 4 −2 1 x 2 −1 4 6 8 10 56 3 48 274 32. y x V Chapter 6 9 9 5 Applications of Integration 0, x 2, x 3 9 8 7 6 5 4 3 2 1 y x 2, y y 9 0 5 y y dy 2 22 d y (2, 5) 5 0 x 123456789 5y 25 y2 2 25 2 2 5 0 25 2 3 34. V 0 y cos x 2 dx 2.4674 36. V 1 ln x 2 d x 3.2332 2 1 x 1 2 5 38. V 0 2 arctan 0.2x 2 dx 15.4115 40. A 3 4 1 3 4 1 2 1 4 b y d 42. V a A x dx or V c A y dy Matches (b) x 1 4 1 2 3 4 1 44. (a) 4 2 2 4 x −4 z (b) 4 2 z (c) 4 4 z 8 y x 8 8 y 8 x 16 y a < c < b. 46. R x V r x, r x h h 0 0 y r2 2 x dx h2 h 0 r y= r x h (h, r) r2 3 x 3h 2 r h3 3h 2 2 x h 12 rh 3 S ection 6.2 4 Volume: The Disk Method 4 275 48. x V r2 r y2, R y r2 y2 2 2 r2 dy y2, r y 0 50. (a) V 0 x 2 dx 0 x dx x2 2 4 8 0 Let 0 < c < 4 and set c h r r h 2 2 y dy y3 3 r3 3 r 2h 3r 2h r h x dx 0 x2 2 c 0 c2 2 4. c2 c r 2h h3 3 h3 h3 3 (b) Set 0 8 8 22 ry r3 2r 3 3 3 2r 3 y Thus, when x 2 2 , the solid is divided into two parts of equal volume. c x dx c2 2 8 (one third of the volume). Then 3 16 ,c 3 d 8 , c2 3 4 3 x dx 0 43 . 3 16 3 (two thirds r To find the other value, set of the volume). Then x h d2 2 16 , d2 3 32 ,d 3 32 3 46 . 3 The x-values that divide the solid into three parts of equal volume are x 4 3 3 and x 4 6 3. 22.2, 0 ≤ x ≤ 11.5 11.5 < x ≤ 15 15 y 8 6 52. y V 0.1x 3 2.95, 11.5 2.2x 2 10.9x 0.1x 3 0 2.2x 2 10.9x 22.2 11.5 2 dx 11.5 2.95 2 d x 15 4 2 0.1x 4 4 2.2x 3 3 10.9x 2 2 22.2x 0 2.95 2x 11.5 x 1031.9016 cubic centimeters 54. (a) First find where y b x2 x 2 4 8 12 16 b intersects the parabola: 4 16 x2 4 4b b 4 x2 4 2 2 4 44 b 5 z 24 4 b V 0 4 b dx bx 2 2 dx 2 4 b b 4 x2 4 2 dx 2 x 2 4 y 4 0 4 0 x2 4 2x 2 2x 3 3 128 3 b x4 16 b2 b 2x 4b 2 8b 8bx 32b 16 d x 4 x5 80 64 5 bx 3 6 32 b 3 16x 0 64 4b 2 64 b 3 512 15 —CONTINUED— 276 Chapter 6 Applications of Integration 54. —CONTINUED— 64 b 3 512 15 64 3 64 3 8 8 3 (b) graph of V b 120 4b 2 (c) V b Vb 8 8b 0⇒b 22 3 >0⇒b 8 is a relative minimum. 3 0 0 4 Minimum Volume is 17.87 for b 10 2.67 56. (a) V 0 fx 2 dx a 2 Simpson’s Rule: b V 3 3 (b) f x 6 10 0 2 10, n 2 10 2 2.6 2 2.1 2 4 1.9 2 2.1 4 2.35 4 2.85 2 2 2.9 2 4 2.7 2 2 2.45 2 4 2.2 2 2.3 2 178.405 0.00249x 4 186.83 cm 3 0.0529x 3 0.3314x 2 0.4999x 2.112 0 0 10 10 (c) V 0 f x 2 dx 186.35 cm 3 58. V 1 10 2 3 2 30 m3 60. 3 y (b) A x 1 bh 2 34 1 24 2 x2 x2 dx x3 3 2 2 x2 34 x2 1 −3 −1 x 1 3 2 V 3 2 4 3 4x −3 32 3 3 Base of Cross Section (a) A x 2 24 2 x2 x2 2 4 − x2 2 4 − x2 b2 44 2 24 x2 dx x 3 32 2 V 2 4 − x2 2 4 − x2 4 4x 128 3 2 4 − x2 —CONTINUED— S ection 6.2 60. —CONTINUED— 12 r 2 2 2 Volume: The Disk Method 277 (c) A x (d) A x 4 x2 2 1 bh 2 1 24 2 x2 x 2 dx 4 x2 4x 4 x3 3 2 2 2 2 4 4x x2 x3 3 2 2 x2 32 3 V 2 4 2 x2 dx 16 3 2 V 2 4 4 − x2 2 4 − x2 2 4 − x2 62. The cross sections are squares. By symmetry, we can set up an integral for an eighth of the volume and multiply by 8. Ay r b2 r2 r2 y2 dy y2 2 y x V 8 0 8 r2y 16 3 r 3 13 y 3 r 0 R2 r2 64. V R2 R2 r2 r2 R2 2 0 x2 r2 x3 3 2 r2 dx R R2 R2 R2 R2 r2 x r2 3 2 r2 32 x2 dx R2 0 r2 r 2 2 4 3 R2 3 r2 32 R2 − r 2 R 66. (a) When a When a (b) y A 1 1 1: x 2: x x 1 1 2 y y 2 1 represents a square. 1 represents a circle. y 1 a=2 a=1 x a 1a 1 −1 1 2 x a 1a dx 4 0 1 xa 1a dx −1 To approximate the volume of the solid, form n slices, each of whose area is approximated by the integral above. Then sum the volumes of these n slices. 278 Chapter 6 Applications of Integration Section 6.3 2. p x hx V 2 0 1 Volume: The Shell Method 4. p x x 8 2 x 1 1 x x1 x 0 hx x dx x2 d x 2 x2 2 x3 3 1 0 x2 x4 0 2 4 x2 dx x3 dx x4 4 2 4 x2 V 2 2 0 2 3 4x 2x 2 2 8 0 6. p x hx V 2 x 12 x 2 6 0 8. p x hx V 2 x 4 2 10. p x x 2 x 4 2 hx x3 dx 14 x 4 2 2x x4 0 2 13 x dx 2 324 0 4x 0 V 8 2 2 0 2x d x 2x 2 d x 23 x 3 2 0 x 4 y 18 15 12 9 6 3 −1 −3 46 2 2x 2 y 0 4x 2x 2 y 2 3 4 3 16 3 2 1 x 1 2 2 1 x 1 2 3 x 1 2 3 4 5 6 −2 −1 −1 12. p x hx V 2 x sin x x x 0 3 2 y sin x dx x 2 cos x 0 1 x π 4 π 2 3π 4 π 2 0 sin x d x 4 −1 14. p y hy V 2 4 y 2 0 p y ≥ 0 on y y2 2 y dy y2 dy 0 2 2, 0 y 16. p y hy V 2 y 16 4 y2 4 y y 16 0 4 y2 dy y3 dy y4 4 64 4 0 3 2 1 x 2 0 2 2 2y y2 y3 3 2 0 16y 8y2 128 −1 −2 −3 −4 4 8 12 2 8 3 2 2 128 S ection 6.3 18. p x hx V 2 0 2 Volume: The Shell Method 279 2 4x 2 x x 2 8x 0 2 20. p x x 2 6 x 4 x 4x 2x 2 hx V 2 x 4x 8x 2 83 x 3 2x 2 d x 2x 3 d x 14 x 2 2 0 6 0 4 x x dx x3 2 2 2 y 4 3 2 0 6x 1 2 4x3 2 dx 192 5 4x2 16 3 y 4 3 2 2 25 x 5 4 2 0 2 1 1 x 1 −1 x 1 3 2 3 4 5 −1 −2 22. (a) Disk Rx V 1 (b) Shell 10 ,r x x2 5 y 0 10 Rx V 2 x, r x 5 0 10 dx x2 10 x2 5 2 8 dx 4 6 4 2 x 1 5 100 1 x x dx −1 x −2 1 2 3 4 5 20 1 1 dx x 5 35 100 100 3 (c) Disk Rx V 1 3 20 1 ln x 1 20 ln 5 1 125 1 496 15 10, r x 5 10 10 200 x 5 1 10 x2 10 x2 2 102 100 3x3 dx 1904 15 y 24. (a) Disk Rx rx a (b) Same as part a by symmetry y a2 3 0 a2 3 a a x2 3 32 (0, a) (0, a) (a, 0) V 2 0 x2 3 3 dx 3a 4 3x 2 3 9 43 53 ax 5 3a 2 3x 4 3 9 23 73 ax 7 13 a 3 (− a, 0) (a, 0) x x a2 a 2x a3 x2 dx 13 x 3 a 0 (0, − a) 2 2 93 a 5 93 a 7 32 a 3 105 280 26. (a) Chapter 6 z 2 Applications of Integration (b) 2 x 5 z (c) 2 z 10 y 2 x −2 5 y x 5 5 y a<c<b 4 28. 2 0 x x dx 2 30. (a) 1 3 4 1 2 1 4 y represents the volume of the solid generated by revolving the region bounded by y x 2, y 0, and x 4 about the y-axis by using the Shell Method. 2 2 16 0 2y 2 dy 0 4 2 2y 2 dy 1 4 1 2 3 4 x 1 represents this same volume by using the Disk Method. y 1 (b) V 4 3 2 1 x 1 −1 2 3 4 5 2 0 x1 x3 dx 2.3222 Disk Method 32. (a) 4 3 2 1 y 34. y tan x, y 1 0, x 0, x 4 Volume Matches (e) y x 1 2 3 4 2 3 (b) V 2 1 2x dx 1 e1 x 1 19.0162 π 4 π 2 x 36. Total volume of the hemisphere is 1 2 Find x0 such that x0 4 3 r3 2 3 3 3 18 . By the Shell Method, p x x, h x 9 x 2. 6 6 2 0 x0 x9 9 0 x2 dx 12 y x2 2x d x x0 2 1 x 2 9 3 9 x 02 32 x2 3 2 18 0 2 9 3 x 02 3 2 −3 −2 −1 −1 −2 −3 1 2 3 18 9 18 2 3 18 2 3 2.920 1.460. x0 Diameter: 2 9 S ection 6.3 r Volume: The Shell Method 281 38. V 4 r r R x r2 r2 x2 dx 2 x2 dx r 4R r 4 r x r2 r x 2 dx 4R 2 2 r 2R r2 2 22 r 3 x2 32 r b 40. (a) Area region 0 abn abn abn abn x 1 axn dx xn 1 a n1 a 1 bn n 1 n n n 1 1 1 1 1 b 0 (c) Disk Method: b V 2 0 x abn b axn dx xn 1 2a 0 xbn bn 2 x 2 bn 2 abn 2 dx b 0 1 abn 1 1 n n 1 2a 2a xn 2 n2 bn 2 n2 2 abn R1 n (b) lim R1 n n→ n→ 1 n abn 2 n n 2 n1 abn b n→ lim n n R2 n (d) lim R2 n n→ n→ n n2 b2 abn lim n n 2 n n 1 2 lim abn b n→ lim b2 abn , the graph approaches the line x 1. (e) As n → 4 42. (a) V 2 0 x f x dx 40 0 34 5800 50 4 10 45 2 20 40 4 30 20 0 2 20 3 121,475 cubic feet 40 20 40 x 1 x 2 12 x 2 25x 2 0 (b) Top line: y 50 x 0 0 40 40 x 20 50 d x 0 20 1 x⇒y 2 2x 40 1 x 2 20 ⇒ y 80 d x 80x dx 50 2x 80 Bottom line: y 20 V 2 0 20 2 20 x 40 2x 2x 2 2 0 50x dx 20 2 20 2 2 x3 6 26,000 3 2 32,000 3 2x 3 3 40 40x 2 20 2 121,475 cubic feet (Note that Simpson’s Rule is exact for this problem.) 282 Chapter 6 Applications of Integration Section 6.4 2. 1, 2 , 7, 10 (a) d (b) y y s 1 Arc Length and Surfaces of Revolution 4. y 1 2 3 2 2x3 2 3 0, 9 9x dx 9 7 4 x 3 4 3 7 10 2 2 10 y s 3x1 2, 9 1 0 2 1 27 1 4 3 2 9x 2 32 0 dx 5 x 3 7 10 1 2 823 27 1 54.929 6. y y s 32 x 2 x 3 4 , 1, 27 1 1 x1 3 3 2 8. y y x5 10 14 x 2 14 x 2 b 1 6x 3 1 2x 4 12 , 1, 2 2x 4 1 y 14 x 2 14 x 2 2 13 27 1 27 1 dx 1 y 2 x2 1 3 x2 27 dx s dx dx 1 2x 4 2 3 2 x2 3 1 2 11 3x 1 2 32 a 3 27 2 1 2 1 dx 3 2 103 2 22 x 3 3 1 23 28.794 1 dx 2x 4 1 6x 3 2 1 15 x 10 1x e 2 1x e 2 1x e 2 2 779 240 3.246 10. y y 1 y 2 e e x 12. (a) y , 0, 2 2 −3 x2 x 5 2, 2≤x≤1 x 2 e x , 0, 2 −5 s 0 1x e 2 2 2 e x x dx (b) 1 y y 2 2x 1 1 1 4x2 2 2 1 2 ex 0 e e x dx 12 e 2 1 e2 3.627 4x 4x 1 4x2 dx 1x e 2 2 0 L (c) L 5.653 14. (a) y 1 1 5 x ,0 ≤ x ≤ 1 (b) 1 y y 2 1 1 1 1 x 1 1 1 2 (c) L 1.132 y= 1 x+1 −1 −2 2 x 4 L 0 1 1 x 4 dx S ection 6.4 Arc Length and Surfaces of Revolution 283 16. (a) y cos x, 2 2 ≤x≤ y = cos x 2 (b) 1 y y 2 sin x 1 sin 2 x 2 (c) 3.820 −2 2 L 2 1 sin 2 x d x −2 18. (a) y ln x, 1 ≤ x ≤ 5 2 (b) 1 y y 2 1 x 1 5 (c) L 1 x2 1 1 dx x2 4.367 −1 6 L 1 −6 20. (a) x y 36 36 y2 , 0 ≤ y ≤ 3 x 2, 3 3 ≤ x ≤ 6 10 (b) dx dy 1 36 2 y 36 3 y2 12 2y (c) L 3.142 ! y2 1 y2 36 y2 dy y2 dy L −10 0 10 3 −5 0 6 36 Alternatively, you can convert to a function of x. y y L 3 3 36 dy dx 6 x2 x 36 1 x2 x2 36 6 dx x2 3 3 6 36 x2 dx 3.142. Although this integral is undefined at x 4 2 0, a graphing utility still gives L y 22. 0 1 s 1 d tan x dx dx 2 y = tan x 1 ( π , 1( 4 π 4 3π 8 x Matches (e) (0, 0) 24. f x (a) d (b) d x2 4 1 4 2, 0, 4 0 0 2 2 144 9 16 16 2 2 128.062 2 1 2 0 9 2 3 2 2 25 0 2 4 3 2 144 25 2 160.151 4 (c) s 0 1 4x x 2 4 2 dx 159.087 (d) 160.287 284 Chapter 6 Applications of Integration 1 and L1 x dx dy e 26. Let y ln x, 1 ≤ x ≤ e, y 1 1 1 d x. x2 1 1 Equivalently, x e y, 0 ≤ y ≤ 1, e y, and L 2 0 1 e 2y d y 0 1 e 2x d x. Numerically, both integrals yield L 10 e x 20 20 x 20 2.0035 28. y y 1 y 2 31 e x 20 1x e 2 1 20 e 10 1x e 4 2 20 e e x 20 x 10 1x e 2 2 2 20 e x 20 s 20 1x e 2 ex 20 20 x 20 dx 20 1 2 20 e dx 10 e x 20 e x 20 20 20 e 1 e 47 ft Thus, there are 100 47 30. y y s 299.2239 4700 square feet of roofing on the barn. 693.8597 68.7672 cosh 0.0100333x 0.6899619478 sinh 0.0100333x 299.2239 1 0.6899619478 sinh 0.0100333x 2 d x 100 or a graphing utility.) 1480 (Use Simpson’s Rule with n x2 x 25 25 25 4 32. y y 1 y 2 25 34. y y S 25 2x 1 x 2 4 9 x2 x2 25 5 25 x 5 4 5 s x 4 3 , 4, 9 9 2x x 4 1 1 dx 9 32 4 1 dx x s 3 4 3 x 2 dx 4 8 3 x dx 2 1 2 5 arcsin 5 arcsin 1 2 4 5 7.8540 8 103 3 3 5 7.8540 53 2 171.258 arcsin Section 6.4 x 2 1 2 5 , 0, 6 4 6 Arc Length and Surfaces of Revolution 285 36. y y 1 y 2 38. y y S 9 2x x2, 0, 3 3 2 0 3 x1 1 0 4x2 dx 12 S 2 0 x 2 5 6 5 dx 4 x2 0 4 6 4x2 4x2 2 8x dx 3 1 373 32 0 2 8 95 6 1 117.319 40. y y 1 y 2 ln x 1 x x2 x2 e 42. The precalculus formula is the distance formula between two points. The representative element is xi 1 , 1, e x2 x2 x2 1 2 yi 2 1 yi xi 2 xi. S 2 1 e x 1 dx 22.943 2 1 dx 44. The surface of revolution given by f1 will be larger. r x is larger for f1. 46. y r2 y 1 y 2 x2 x r2 r2 r2 x2 r 48. From Exercise 47 we have: a x2 S 2 0 a rx dx r2 x2 2x d x r2 x2 a r 0 S 2 r r r2 r dx r x2 2 rx r2 r2 r x2 dx 2r 2 2r 2r r2 2r x2 0 2 4 r2 r r2 r2 a2 a2 r 2 r h where h is the height of the zone y r h r−h x x2 + y2= r2 286 Chapter 6 Applications of Integration 50. (a) We approximate the volume by summing 6 disks of thickness 3 and circumference Ci equal to the average of the given circumferences: 6 6 V i 1 ri 2 3 i 1 2 Ci 2 2 3 3 4 70 2 2 6 Ci 2 i 1 3 4 50 2 65.5 65.5 70 2 66 2 66 2 58 2 58 2 51 2 51 2 48 2 3 57.75 2 4 67.75 2 68 2 62 2 54.5 2 49.5 2 3 21813.625 5207.62 cubic inches 4 (b) The lateral surface area of a frustum of a right circular cone is s R S1 32 50 2 65.5 2 65.5 9 50 2 12 r . For the first frustum, r s 50 2 50 65.5 2 2 12 65.5 2 . h Adding the six frustums together, S 50 2 70 2 58 2 224.30 1168.64 (c) r 0.00401 y 3 20 R 2 12 65.5 66 51 9 9 9 15.5 2 4 2 7 2 65.5 2 66 2 51 2 202.06 58 48 70 9 8 2 3 2 4.5 2 2 12 2 12 2 12 9 9 2 12 2 12 208.96 208.54 174.41 150.37 18 0.1416 y 2 1.232 y 7.943 (d) V 0 18 r2 dy 2 ry 0 5275.9 cubic inches 1 r y 2 dy S 1179.5 square inches −1 −1 19 52. Individual project, see Exercise 50, 51. x2 9 y2 4 x2 ,0 ≤ x ≤ 3 9 x2 9 12 54. (a) 1 2 1 4 (b) y 1 x2 9 x2 9 y 2 2 1 2 1 1 2x Ellipse: y1 y2 2 2x 9 2x x2 9 3 1 x2 9 1 39 −5 2 5 L 0 81 4x2 dx 9x2 x2 + y = 1 94 −4 (c) You cannot evaluate this definite integral, since the integrand is not defined at x same reason. Also, the integrand does not have an elementary antiderivative. 56. Essay 3. Simpson’s Rule will not work for the Section 6.5 Work 287 Section 6.5 2. W Fd Work 11,200 ft lb 4. W Fd 9 2000 1 2 2800 4 5280 47,520,000 ft lb b 6. W a F x dx is the work done by a force F moving an object along a straight line from x a to x b. 9 10 8. (a) W 0 7 6 dx 20 dx 0 54 ft 9 lbs 10x 90 dx 140 20 10. W 0 5 x dx 4 lb 52 x 8 3.33 ft 10 6 (b) W 40 in lb 7 160 ft 9 lbs x3 81 9 (c) W 0 9 12 x dx 27 x dx 9 ft 0 9 0 lbs 18 ft lbs (d) W 0 2 32 x 3 2 27 3 12. F x 800 W kx k 70 ⇒ k 70 14. F x 80 7 70 kx k1 4 15 40x 2 7 70 0 k 4 F x dx 0 0 80 x dx 7 W 2 0 15x d x lb 15x 2 0 240 ft 28000 n cm 16 280 Nm kx 2 2 270x 2 16 h→ 16 0 5 24 h h 4000 16. W W 7.5 0 5 24 kx dx 540x d x k ⇒k 72 4.21875 ft 540 lbs 18. W 80,000,000 dx x2 4000 80,000,000 x 80,000,000 h 20,000 lb 16 lim W 20,000 mi ton 2.1 1011 ft 20. Weight on surface of moon: 1 12 6 2 tons Weight varies inversely as the square of distance from the center of the moon. Therefore, Fx 2 k W 1100 k x2 k 1100 2.42 1150 2 10 6 2.42 x2 10 6 dx 2.42 x 10 6 1150 2.42 1100 10 6 1 1100 ton 1 1150 1.01 10 9 ft lb 95.652 mi 288 Chapter 6 Applications of Integration 22. The bottom half had to be pumped a greater distance then the top half. 24. Volume of disk: 4 y y Weight of disk: 9800 4 Distance the disk of water is moved: y 12 W 10 y 9800 4 dy 39,200 39,200 y2 2 22 12 10 862,400 newton–meters 2 y 3 2 y 26. Volume of disk: y 7 Weight of disk: 62.4 Distance: y (a) W (b) W 4 62.4 9 4 62.4 9 2 2 y 3 2 5 y 4 3 2 y x y3 0 6 dy 4 62.4 9 4 62.4 9 14 y 4 14 y 4 2 −4 −3 −2 −1 1 2 3 4 110.9 ft 0 6 lb lb y3 dy 4 7210.7 ft 4 28. Volume of each layer: y 3 3 3 3 y y y 3 y 4 y Weight of each layer: 55.6 y Distance: 6 3 (2, 3) 3 y 3 2 W 0 55.6 6 yy 3 dy 55.6 0 18 3y 3y2 2 lb y2 dy y3 3 3 0 1 y = 3x − 3 x 1 2 3 4 55.6 18y 3252.6 ft 30. Volume of layer: V Weight of layer: W Distance: 2.5 12 2 42 24 25 4 25 4 y2 y y2 y Ground level 12 y 19 2 6 y 25 4 2.5 2.5 3 −9 −6 −3 −3 −6 3 19 2 −y x W 2.5 42 24 1008 19 2 y2 25 4 19 2 y2 dy 6 9 y dy 2.5 2.5 25 4 y2 y dy The second integral is zero since the integrand is odd and the limits of integration are symmetric to the origin. The first integral represents the area of a semicircle of radius 5 . Thus, the work is 2 W 1008 19 2 5 2 2 1 2 29,925 ft lb 94,012.16 ft lb. Section 6.5 32. The lower 10 feet of chain are raised 5 feet with a constant force. W1 3 10 5 150 ft lb Work 289 The top 5 feet will be raised with variable force. Weight of section: 3 y Distance: 5 W2 W 3 0 y 5 5 W2 y dy 150 3 5 2 75 2 5 y 2 0 75 ft 2 lb W1 375 ft 2 lb 34. The work required to lift the chain is 337.5 ft lb (from Exercise 31). The work required to lift the 500-pound load is W 500 15 7500. The work required to lift the chain with a 100-pound load attached is W 337.5 6 7500 7837.5 ft 3 12 4 lbs 6 36. W 3 0 12 2y d y 2y 2 0 3 12 4 2 108 ft lb 38. Work to pull up the ball: W1 500 40 20,000 ft lb 40. p 2500 W k V k ⇒k 1 3 1 Work to pull up the cable: force is variable Weight per section: 1 y Distance: 40 40 2500 3 x 40 x dx lb W2 8000 20,000 2 800 20,800 ft lbs 4 15,000 2 10,000 lb 1 40 2 40 2500 dV V 2500 ln V 1 W2 0 x 2 0 2500 ln 3 2746.53 ft lb 800 ft W 42. (a) W (b) W W1 FD 16,000 ft 2 22,000 20 0 36 24,88.889 ft 4 20,000 lb 4 5000 0 (c) F x 25,000 16,261.36x 4 85,295.45x 3 157,738.64x 2 104,386.36x 32.4675 0 0 2 (d) F x (e) W 0 0 when x 2 0.524 feet. F x is a maximum when x 25,180.5 ft lbs 0.524 feet. F x dx 4 44. W 0 ex 1 dx 100 2 2 11,494 ft lb 46. W 0 1000 sinh x d x 2762.2 ft lb 290 Chapter 6 Applications of Integration Section 6.6 2. x 7 3 7 4 4 Moments, Centers of Mass, and Centroids 2 3 35 8 86 17 11 4. x 12 6 14 12 1 6 6 2 3 30 11 11 8 0 6. The center of mass is translated k units as well. 8. 200x 200x 750x x 550 5 2750 2750 32 feet 3 x (Person on left) 550x 10. x y x, y 10 1 10 0, 0 25 10 2 1 10 5 5 4 12 2 0 0 12 3 y 6 12 65 12 1 6 12. x 25 50 25 15 6 15 2 2 15 15 2 15 8 2 15 2 y 8 93 40.5 62 27 15 15 2 y 6 m2 (5, 5) 96 40.5 64 27 8 6 4 x, y x 6 m3 (− 4, 0) −4 −2 2 62 64 , 27 27 6 m3 (6, 8) −2 m1 4 (1, −1) m2 (− 1, 5) 2 −2 −2 m1 (2, 3) x 2 m4 (2, − 2) 6 8 2 14. m 0 2 12 x dx 2 1 12 x 22 4 5 4 3 x3 6 2 0 4 3 2 2 y Mx 0 12 x dx 2 3 5 1 2 2 8 x4 dx 0 40 x5 0 32 40 4 5 2 1 ( 3, 3) 25 x 1 2 y Mx m 2 My 0 x My m 33 , 25 12 x dx 2 2 4 3 3 2 2 x3 dx 0 8 x4 0 2 x x, y Section 6.6 1 Moments, Centers of Mass, and Centroids 291 16. m 0 1 x x dx x 1 2 2 32 x 3 x dx 2 x2 2 1 1 y 0 6 1 Mx 0 xx 2 6 12 x 0 x2 dx x2 22 x3 3 1 0 12 3 4 1 2 y My Mx m 1 ( x, y ) 1 4 1 x 0 x 6 15 x dx 0 x3 2 x2 dx 2 52 x 5 x3 3 1 0 x 15 1 4 1 2 3 4 1 x x, y My m 21 , 52 9 2 5 18. m 0 x 23 x 3 9 2 1 x2 6 1 2 9 1 x 3 18 9 1 27 2 dx 0 x 9 2 1 x 3 2x 27 9 1 x dx 3 0 x Mx 0 9 1 x 3 1 x 13 x 3 9 2 0 2 9 1 1 dx 2 x dx 3 36 45 3 2 2 x 0 9 0 1 x 3 12 x 9 2 x 1 x dx 3 2 x 0 13 x 3 x3 27 x 81 2 12 x 9 27 22 1 dx 1 x 3 2 x dx x2 26 9 43 x 3 1 My 0 x 486 5 My m 81 5 9 2 18 5 , 52 y 5 4 3 2 1 1 x 3 81 5 Mx m x3 0 2 12 x dx 3 25 x 5 2 13 x 9 9 0 x 18 ;y 5 45 4 9 2 5 2 x, y (9, 4) (18, 5 ) 52 (0, 1) x 2 4 6 8 10 −2 −1 292 Chapter 6 8 Applications of Integration 3 53 x 5 8 0 20. m 2 0 4 x2 3 dx 0. 2 4x 128 5 12 8 y By symmetry, My and x 8 ( x, y ) Mx y x, y 2 0 4 5 128 x2 3 4 2 20 7 x 23 dx 16x 3 73 x 7 8 0 512 7 −8 −4 −4 x 4 8 512 7 0, 20 7 2 22. m 0 2 2y 2y 2 y2 dy y2 2y 2 5 y2 dy 1 y2 y2 dy y3 3 2 0 4 3 y4 y5 5 2 0 y My 0 4y 3 23 8 15 2 ( x, y ) 1 x Mx My m 2 83 15 4 y 2y 0 2y 3 3 y4 4 2 0 4 3 x 1 2 y x, y Mx m 2 ,1 5 2 43 34 24. m 1 2 y y 2 2 2 2 2 9 2 y2 2 y2 dy y2 y 2 y2 2 2y y2 dy y 2 3 3 y3 3 2 1 9 2 3 2 1 y My 1 2 2 x Mx My m y 1 y4 dy 8 5 y2 dy y3 dy 1 2 2 y5 5 2 1 36 5 ( x, y ) x 1 2 3 4 36 5 2 −1 yy 1 2 2y 1 y2 y3 3 y4 4 2 1 9 4 y x, y Mx m 81 , 52 4 92 49 26. A 1 1 dx x 4 1 4 ln x 1 ln 4 1 x 4 4 1 Mx My 1 2 4 1 dx x2 1 dx x 1 2 x 1 1 8 1 2 3 8 x 1 3 Section 6.6 2 2 Moments, Centers of Mass, and Centroids 16 3 32 3 293 28. A 2 x2 1 2 2 4 dx 44 8x 3 3 2 0 4 x2 dx 1 2 32 5 2 8x x4 8x 2 32 2x 3 3 2 16 0 Mx x2 2 x 2 dx 2 16 d x 256 15 2 1 x5 25 My 16x 2 64 3 0 by symmetry. 4 3 30. m 0 4 xe xe x2 dx xe dx 2.3760 4 x2 ( x, y ) x2 Mx 0 4 2 x 2e x2 dx 2 x 2e x 0 dx 0.7619 −1 −1 5 My 0 5.1732 x y My m Mx m 2.2 0.3 Therefore, the centroid is 2.2, 0.3 . 2 32. m 2 2 8 x2 4 dx 6.2832 8 x2 4 2 3 Mx y x 34. Mx m 1 8 2 4 22 x 0.8 dx 32 2 1 x2 4 2 dx 5.14149 −3 −1 3 0 by symmetry. Therefore, the centroid is 0, 0.8 . A 1 A x bh 1 ac 11 ac 2 1 2ac c 0 c 0 ac y y= c x b b y c 2ab y c 2 a a2 dy c b y c 2 (b, c) (a + b, c) dy ( x, y ) y = c (x − a ) b x (0, 0) (a, 0) 1 ab 2 y 2ac c y x, y 1 ac b 2 c a 2y 0 1 abc 2ac b y c dy a 2c 1 y2 c2 1 b 2 c 0 a c 2 y 0 b y c a ac , 2 This is the point of intersection of the diagonals. 294 36. x Chapter 6 Applications of Integration y 0 by symmetry A 1 A y 12 r 2 2 r2 21 r2 2 r r r2 r x2 2 r x dx 4r 3 1 r 2x r2 x, y 0, 4r 3 x3 3 r r 1 4r3 r2 3 1 38. A 0 1 3 1 2x x2 dx 1 3 1 A x y 1 3 0 1 x1 1 2x 2x 2 4x 2 x2 dx x2 1 3 0 x 2x 2 3 2 43 x 3 x3 dx 1 3 2x x5 5 1 0 x2 2 x2 7 10 23 x 3 2 x4 4 1 0 1 4 3 0 2x x4 dx x2 dx 3 x 2 1 0 dx 3 2 x, y 1 1 0 4x 3 x4 17 , 4 10 0 by symmetry 2n 40. (a) M y b (b) y > x 2n d x 0 (d) 2n My 2n xb b b because there is more area above y 2 than below. b 2 because bx 2n x 2n b 1 is an odd function. x2n 2n n y 1 3 5b 2 5 9b 3 7 13 b 4 9 17 b b b (c) M x 2n x2n b 2 x 4n 1 4n 1 b 4n 4n x 2n 1 dx b 2n 2n 1 2n 1 2n dx 12 b 2 2n b b x 4n dx 12 bx 2 b 2 b 1 2n 2n b 2n (e) lim y b n→ n→ lim 2n 4n 1 b 1 1 b 2 1 2n 4n b 4n 4n 1 2 bx 1 2n 1 2n (f) As n → b , the figure gets narrower. y b A 2n b b x 2n 1 2n 1 2n y = x 2n 0 1 2n 2b y Mx A b 1 2n 4n b 4n 4n b 24n 1 4n 2n 4n b 2n 2n 1 1 1 2n 4n y=b x 1 b 1 − 2n b 2n b Section 6.6 42. Let f x be the top curve, given by l x f d 0 2.0 0.50 0.5 1.93 0.48 2 Moments, Centers of Mass, and Centroids 295 d. The bottom curve is d x . 2.0 0 0 1.0 1.73 0.43 1.5 1.32 0.33 (a) Area 2 0 fx 2 1.50 34 d x dx 4 1.45 4.62 2 1.30 4 .99 0 (b) f x dx (c) x, y 0.1061x 4 0.06126x 2 1.9527 .4862 0.02648x 4 y Mx A 0.01497x 2 1.068 2 1 13.86 3 2 4.9133 4.59998 0, 1.068 Mx 2 2 fx 2 fx 2 dx fx 2 d x dx dx dx 2 2.808 4 1.6335 0 0 2 3.75 34 1 29.878 6 y x, y Mx A 4 3.4945 4.9797 3 f d −2 2 0 4.9797 4.62 1.078 0, 1.078 13 1 7 , , 2, , and , 1 22 2 2 27 2 21 25 2 7 15 2 7 25 14 15 14 y 4 3 44. Centroids of the given regions: Area: A x y x, y 3 2 31 2 33 2 25 15 , 14 14 7 ,P 41 7 8 7 16 2 7 22 7 21 2 7 2 1 m1 m2 m3 x 1 2 3 4 46. m1 m2 7 2 8 7 6 8 0, y 287 ,P 64 2 0. 0, 55 16 5 4 3 m2 By symmetry, x y x, y 2 7 4 7 16 74 0, 5523 2128 287 64 55 16 287 64 0, 2.595 16,569 6384 5523 2128 m1 3 −2 −1 1 −2 x 1 2 1 3 2 296 Chapter 6 Applications of Integration 50. V 2 rA 2 34 24 2 48. Centroids of the given regions: 3, 0 and 1, 0 Mass: 8 y x x, y 0 81 8 8 8 3 3 8 8 3 ,0 1.56, 0 6 52. A 2 6 2x x2 x 2 2 dx 2 dx 2, x 2 32 3 4 x 3 6 6 2 32 2 32 3 6 y My Let u 2 2 xx 2, du 4 2 dx 4 (6, 4) x 4 u dx: 2 2 My 2 0 u 64 5 u du 704 15 22 5 2 0 u3 2u1 2 du 2 25 u 5 2 43 u 3 4 2 0 2 ( x, y ) x 4 6 2 x r V My A x 704 15 32 3 22 5 2 22 5 2 rA 32 3 1408 15 294.89 54. A planar lamina is a thin flat plate of constant density. The center of mass x, y is the balancing point on the lamina. 56. Let R be a region in a plane and let L be a line such that L does not intersect the interior of R. If r is the distance between the centroid of R and L, then the volume V of the solid of revolution formed by revolving R about L is V 2 rA where A is the area of R. 58. The centroid of the circle is 1, 0 . The distance traveled by the centroid is 2 . The arc length of the circle is also 2 . 22 4 2. Therefore, S y 2 1 C d x 1 3 −1 −1 −2 Section 6.7 Fluid Pressure and Fluid Force 297 Section 6.7 2. F PA Fluid Pressure and Fluid Force 4992 lb 4. F 62.4 h 4 48 62.4 h 48 62.4 5 16 62.4 4 48 6. h y Ly F 3 4 y 3 3 11,980.8 lb y 8. h y Ly 4 y y dy 3 3y 0 y 24 0 y2 y2 4 0 32 2 F 62.4 2 y2 dy 332.8 lb 62.4 0 3 3 4 62.4 3 4 62.4 3 62.4 y2 dy y 2 4 3 y2 3y 2 2 y3 3 3 0 374.4 lb −1 1 x Force is one-third that of Exercise 5. y 4 1 −3 2 1 x −2 −1 1 2 10. h y Ly F 4 3 62.4 y 9 0 y x y2 y 4 3 9 3 0 −1 −1 −2 1 9 y2 y2 dy 12 −4 3 2 62.4 3 0 2y d y 748.8 lb 4 62.4 9 9 12. h y L1 y L2 y F 1 2y 32 y2 32 3 y 1+ 3 2 32 32 2 3 2 y lower part y 3 22 23 2 2 9800 0 upper part 1 32 y3 3 3 9 0 3 y y dy 3 22 22 1 32 y3 3 y3 2 62 2 1 y dy 3 2 −3 −2 −1 1 3 x 2 3 19,600 y2 2 3 2y 1 3 2y 2 4 1 18y y 3 22 19,600 92 2 4 44,100 3 2 2 Newtons 298 14. h y Ly F Chapter 6 6 1 5 Applications of Integration 16. h y Ly 2 140.7 3 y y 4 3 0 9 y2 y2 0 9800 0 16 y 2 y dy 25 F 171,500 Newtons 9800 6y y 6 4 3 9 9 y2 y2 32 y2 dy 2y d y 0 0 140.7 4 3 140.7 4 3 y 2 x 3 2 9 3 3376.8 lb 3 −3 −2 −1 1 2 3 x −2 2 −4 −6 18. h y Ly F 5 y 5 y 3 0 1 y x −1 2 3 4 6 140.7 3 0 y5 5y 3 5 y dy 3 52 y dy 3 −2 −3 −4 −5 (5, −3) 140.7 140.7 140.7 52 y 2 45 2 15 53 y 9 0 3 1055.25 lb 20. h y Ly F 42 3 2 2 1 2 32 y 9 3 2 4y 2 32 y 9 4y 2 d y 63 32 9 4y 2 d y 32 21 4 32 9 32 4y 2 8y d y The second integral is zero since it is an odd function and the limits of integration are symmetric to the origin. 3 The first integral is twice the area of a semicircle of radius 2 . 9 4y 2 2 9 4 94 y2 141.75 445.32 lb. 11 35 2 17 5 10 5 Thus, the force is 63 22. (a) F (b) F wk r2 wk r2 62.4 7 62.4 5 22 32 1747.2 lbs 2808 lbs 24. (a) F (b) F wkhb wkhb 62.4 62.4 5148 lbs 10,608 lbs Review Exercises for Chapter 6 26. From Exercise 21: F 64 15 1 2 2 299 753.98 lb 28. h y 3 y 5x 2 x 2 4y 5 2 4y 5 y 3 30. h y 4 for x, you obtain y. Ly F 4y 5 y y 5 y y 12 2 62.4 0 y 7 16 2 4 Solving y x Ly F y2 7 16 7 16 y 16 y2 12 4 y 12 y2 dy y2 dy 21373.7 lb 62.4 2 0 3 3 3 0 y dy dy 546.265 lb 62.4 7 0 y 2 124.8 y 5 4 10 8 6 x 2 1 x −3 −2 −1 −1 1 2 3 −6 −4 −2 2 4 6 32. Fluid pressure is the force per unit of area exerted by a fluid over the surface of a body. 34. The left window experiences the greater fluid force because its centroid is lower. Review Exercises for Chapter 6 5 2. A 12 4 4x y 1 dx x2 1 x 5 12 1 4. A 0 1 y2 y2 0 1 2y 2y 1 2 dy 1 dy 1 dy 81 5 6 y 0 1 ( 2 , 4) 5 (5, 4) y 3 1 31 0 3 2 1 1 2 3 4 1 ( 5, 25 ) 6 1 3 y x 3 2 (−1, 1) 1 1 2 −2 3 −2 1 −2 x 300 Chapter 6 2 Applications of Integration 2 6. A 1 2 y 2 1 3 y 12 y 2 y2 y2 dy 13 y 3 2 1 1 dy 8. A 2 6 2 csc x d x 2 2 2x 9 2 2 2 2 3 ln csc x 0 ln 2 cot x 6 2y y 3 3 3 ln 2 3 1.555 (5, 2) 2 1 x 2 −1 −2 3 4 5 3 y (π , 2) 6 1 ( 56π, 2) (2, −1) π 6 π 3 π 2 2π 3 5π 6 x 5 3 3 10. A y 2 3 1 2 sin y 7 3 cos y d y 5 5 3 3 cos y y 2 7 5 3 3 1 dy 2 12. Point of intersection is given by: x3 A x2 0.783 4x 3 3 4x 0⇒x x2 0.783. sin y 3 x3 dx 14 x 4 0.783 0 23 y 3 0 3x 2x 2 13 x 3 2 ( 21 , 73π ) ( 21 , 53π ) 1 ( 2, π ) 3 1.189 4 −1 x 2 3 − 2 (.7832, .4804) −2 −1 2 14. A 2 0 2 2x 2 4x 2 0 x4 x4 dx 15 x 5 2 0 2x 2 d x 16. y y x x 1 2 2 1⇒x ⇒x 1 1 1 32 y2 2y y2 x 2 1 x 4 1 1 1 2 4 2 x3 3 10 128 15 8.5333 A 0 5 2y x 1 dy dx 5 (−2, 8) (2, 8) 1 2 x 3 −4 −2 4 y 1 2 1 4 3 (0, 0) 3 (5, 2) 2 1 x (1, 0) 2 −1 −2 3 4 5 R eview Exercises for Chapter 6 1 5 301 18. A 0 2 dx 1 2 x 1 dx 5 y x A y2 2 1 y2 1 dy 2 4 3 (5, 2) 0 13 y 3 y 0 14 3 1 (1, 0) x 2 3 4 5 20. (a) R1 t 40 5.2834 1.2701 t 5.2834 e0.2391t (b) R2 t Difference 10 15 5.28 e0.2t R1 t 10 R2 t dt 171.25 billion dollars 0 0 10 22. (a) Shell 2 2 (b) Shell 2 V y 4 3 2 0 y3 dy 2 y4 0 8 V 2 0 2 2 2y 2 0 y y2 dy y3 dy 14 y 4 2 0 2 2 y 23 y 3 8 3 2 1 4 x 1 2 3 4 3 1 x 1 2 3 4 (c) Disk 2 2 (d) Disk y4 dy 0 V y 5 y5 0 32 5 2 V 0 2 y2 y4 0 1 2 12 d y 2y 2 d y 23 y 3 2 0 4 3 15 y 5 y 176 15 2 1 x 1 2 3 4 3 2 1 x 1 2 3 4 5 4 302 Chapter 6 Applications of Integration (b) Disk 24. (a) Shell a V 4 0 x 2b a b a a a a2 a2 x2 x2 x2 dx 12 V 2x d x 42 ab 3 2 0 b2 2 a a2 x2 dx 13 x 3 a 0 0 a 32 0 2 b2 2 ax a2 4 ab2 3 y 4b 2 a 3a y (0, b) 2 x2 + y =1 a2 b2 (0, b) 2 x2 + y =1 a2 b2 x x (a, 0) (a, 0) 26. Disk 1 28. Disk 2 0 V 1 1 1 0 2 1 x2 dx V 0 1 e e 0 x2 dx 1 2 arctan x 2 2 2x dx 1 2 2 e 2x 0 4 0 y 2e 2 2 1 e2 2 y 3 2 x 1 1 −2 −1 −1 x 1 2 30. (a) Disk 0 (b) Shell x2 x 1 0 V 1 dx x2 0 1 u x u2 x 1 1 y x3 1 dx dx V 2u du 0 −1 x x4 4 y x3 3 12 2 1 1 x2 x u2 0 1 1 dx −1 4 x 1 2 u 2du 2u 4 25 u 5 u2 du 13 u 3 1 0 −1 4 0 u6 17 u 7 −1 4 32 105 Review Exercises for Chapter 6 1 bh 2 3 a2 a 303 32. A x 1 2 a2 2 x2 x2 3 a2 x2 34. y y x3 6 12 x 2 12 x 2 3 1 2x 1 2x 2 1 2x 2 2 V 3 a a2 4a3 3 x2 dx 3 a 2x x3 3 a a 1 y 2 3 s 1 12 x 2 1 dx 2x 2 13 x 6 1 2x 3 1 14 3 Since 4 3 a 3 3 a 3 10, we have a 3 1.630 meters. 5 3 2. Thus, 53 2 2 a 2− x 2 2 a 2− x 2 2 a 2− x 2 tan x has f x sec 2 x, this integral 36. Since f x represents the length of the graph of tan x from x 0 to 4. This length is a little over 1 unit. Answers (b). x 38. y 2x y 1 x 1 2 0 1 y 2 1 x 3 x x 2x 1 x x 3 S 1 3 dx 56 3 4 0 x 1 dx 4 2 x 3 1 32 0 40. F 50 F W kx k9 ⇒ k 50 x 9 9 0 42. We know that 50 9 dV dt V 25 2 x 9 9 0 4 gal min 12 gal min 7.481 gal ft3 r 2h dh 9 dt 9 dV dt 9 3.064 t 8 7.481 150 1 h 9 8 ft3 min 7.481 50 x dx 9 lb dV dt lb dh dt 225 in 18.75 ft 3.064 ft min. Depth of water: Time to drain well: t 49 12 150 49 minutes 3.064 588 gallons pumped Volume of water pumped in Exercise 41: 391.7 gallons 391.7 52 x Work 588 x 588 52 391.7 78 ft 78 ton 304 Chapter 6 Applications of Integration 44. (a) Weight of section of cable: 4 x Distance: 200 200 x 200 W 4 0 200 x dx 2 200 x 2 0 80,000 ft lb 40 ft lb ton 30 ft ton (b) Work to move 300 pounds 200 feet vertically: 200 300 Total work work for drawing up the cable 40 ft b 60,000 ft work of lifting the load ton ton 30 ft ton 70 ft 46. W a F x dx 2 9x 4 3x 9 Fx W 0 6, 16, 6 dx 0≤x≤9 9 ≤ x ≤ 12 12 9 2 x 9 12 x 9 9 3 4 x 3 16x 9 16 d x 12 9 6x 0 22 x 3 96 192 54 54 144 13 x 3 51 ft 3 1 lbs 32 3 48. A 1 2x 3 32 3 32 3 3 x2 dx x2 3x 1 A x y x 2x 1 3 3 2x x 2 dx 3 43 x 3 2 3 32 3 3x 1 2x 2 3 x 3 dx 12x 4x 2 3 32 x 32 2 x4 dx 23 x 3 14 x 4 3 1 1 y 31 32 2 3 9x 64 x4 dx 15 x 5 3 1 1 3 64 17 5 9 1 9 6x 2 6 ( x, y ) 3 x, y 17 1, 5 −3 x 3 6 8 50. A 0 x2 3 5 16 5 16 8 1 x dx 2 3 53 x 5 12 x 4 8 0 16 5 y 6 4 2 1 A x ( x, y ) x x2 3 0 1 x dx 2 13 x 6 8 0 x 2 −2 4 6 8 5 3 83 x 16 8 y 51 16 2 15 2 16 x, y 10 40 , 3 21 8 10 3 x4 3 0 12 x dx 4 13 x 12 8 0 3 73 x 7 40 21 ...
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This note was uploaded on 05/18/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

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